Answer:
D) either 436 Hz or 444 Hz
Explanation:
frequency of the tuning fork, F₁ = 440 Hz
frequency of the piano, F₂ = ?
Beat frequency, F = 4 Hz
Beat frequency is given as the difference between the frequency of the two instruments and it is given by;
F = F₂ - F₁ or F = F₁ - F₂
F₂ = F + F₁ or F - F₁ = - F₂
F₂ = 4 Hz + 440 Hz or 4 - 440 = - F₂
F₂ = 444 Hz or - 436 = - F₂
F₂ = 444 Hz or F₂ = 436 Hz
Therefore, the frequency of the piano is 444 Hz or 436 Hz
2. The components of vector A are given as follows:
Ax = 5.6 Ay = -4.7
What is the angle between vector A and positive direction of x-axis?
Answer:
50 degree.
Explanation:
Given that the components of vector A are given as follows: Ax = 5.6 Ay = -4.7
The angle between vector A and B in the positive direction of x-axis will be achieved by using the formula:
Tan Ø = Ay/Ax
Substitute Ay and Ax into the formula above.
Tan Ø = -4.7 / 5.6
Tan Ø = -0.839
Ø = tan^-1(-0. 839)
Ø = - 40 degree
Therefore, the angle between vector A and B positive direction of x-axis will be
90 - 40 = 50 degree.
A magnetic field is created by ____.
A moving electric charges
b elctromagnetic pulses
c a strong current
d a change in the current of a wire
Calculate the length of segment RS with midpoint, M , if RM = 5x and MS = x + 12.
A. 3
B. 25
C. 30
D. 36
E. 15
Answer:
the full segment is: 30 units long
which coincides with answer C in your list
Explanation:
If M is the midpoint, then it divides the segment in two equal parts. Then, we can say that:
RM = MS (equality among the two parts of the divided segment)
replacing RM with "5 x" and MS with "x + 12", we get:
5 x = x + 12
solving for x:
5 x - x = 12
4 x = 12
then x = 3
With this info, we can calculate the length of each half of the segment and consequently its full length:
RM = 5 x = 5 (3) = 15
then the full segment is: 30 units long
what is machinery
............
How does deploying slats or slots on an airfoil affect CL max and Stall AOA?
a) Stall AOA increases, CL max increases.
b) Stall AOA decreases, CL max increases.
c) Stall AOA increases, CL max decreases.
d) Stall AOA decreases, CL max decreases.
Answer: a. Stall AOA increases, CL max increases.
Explanation:
Deploying slats or slots on an airfoil affect CL max and Stall AOA by increasing Stall AOA and also increasing CL max.
It should be noted that the coefficient of lift CL would rise through the use of slots due to increase in boundary later energy. The slots also leads to delay of stall by through increase in AOA.
An object is released from rest from a top of a building 90 meters high. Neglect air friction.
What is the volume of its acceleration?
Calculate the time it takes to reach the floor?
With what velocity does it reach the floor?
How fast is it moving when it is 50 meter above the floor?
Answer:
1. a = 9.8 m/s²
2. t = 4.28 s
3. Vf = 42 m/s
4. Vf = 28 m/s
Explanation:
1.
Since, the body is under free fall motion. Therefore, the value of its acceleration shall be equal to the acceleration due to gravity.
a = 9.8 m/s²
2.
The time taken by the ball to reach the ground can be calculated by using second equation of motion:
h = Vi t + (1/2)gt²
where,
h = height = 90 m
Vi = initial velocity = 0 m/s
t = time taken = ?
Therefore,
90 m = (0 m/s)t + (1/2)(9.8 m/s²)t²
t = √(18.36 s²)
t = 4.28 s
3.
In order to find final velocity we use first equation of motion:
Vf = Vi + gt
Vf = 0 m/s + (9.8 m/s²)(4.28 s)
Vf = 42 m/s
4.
when the ball is at height of 50 m, it means it has covered:
h = 90 m - 50 m = 40 m
we use third equation of motion at this point:
2gh = Vf² - Vi²
(2)(9.8 m/s²)(40 m) = Vf² - (0 m/s)²
Vf = √(784 m²/s²)
Vf = 28 m/s
What does this picture show?
A. Good accuracy, poor precision
B. Poor accuracy, good precision
C. Good accuracy, good precision
D. Poor accuracy, poor precision
sam exerts a force of 65 N on a lawn mower with a mass of 25 kg. which formula can be used to calculate the acceleration of a lawn mower
Answer:
F = ma - with this you will get 2.6 m/s^2
Explanation:
Use this formula to get the acceleration...
where F is force, M is mass and A is acceleration
by using this we get...
65 = 25 * a
so, a = 65/25
Therefore, the acceleration is 2.6 m/s^2
Hope that helped :)
The acceleration of a lawn mower will be 2.6 m/s². It is the ratio of force and the mass.
What is acceleration?The rate of change of velocity with respect to time is known as acceleration. According to Newton's second law, the eventual effect of all forces applied to a body is its acceleration.
The given data in the problem is;
The force act on a lawn mower,[tex]\rm F = 65 \ N[/tex]
Mass of lawn mower,[tex]\rm m = 25 kg[/tex]
The acceleration of a lawn mower is,[tex]\rm a = ?[/tex]
Acceleration, is found as the ratio of force and the mass.
[tex]\rm F= ma \\\\ a = \frac{F}{m} \\\\ a= \frac{65}{25} \\\\ a = 2.6 \ m/s^2[/tex]
Hence, the acceleration of a lawn mower will be 2.6 m/s².
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Using the equation for for Newton's Second law, m=F/a solve the following problem. You have been given an object with a force of 10N and an acceleration of 2 m/s2, what is the mass?
Group of answer choices
1. 8g
2. 3g
3. 20g
4. 5g
Answer:
4. 5g
Explanation:
F=ma so, m=Fa. All you have to do is 10/2. Don't be confused by the units. Mass will normally be in grams.
What were the physical activities in your childhood that you still do today? Do you spend more time now in doing these activities as compared before?
A sprinter practicing for the 200-m dash accelerates uniformly from rest at A and reaches a top speed of 41 km/h at the 61-m mark. He then maintains this speed for the next 83 meters before uniformly slowing to a final speed of 34 km/h at the finish line. Determine the maximum horizontal acceleration which the sprinter experiences during the run. Where does this maximum acceleration value occur
Answer:
2.13 m/s^2.
Find the remaining answer in the explanation
Explanation:
Given that a sprinter practicing for the 200-m dash accelerates uniformly from rest at A and reaches a top speed of 41 km/h at the 61-m mark. He then maintains this speed for the next 83 meters before uniformly slowing to a final speed of 34 km/h at the finish line.
1.) The maximum horizontal acceleration will be when she attained her maximum speed. That is,
Speed = 41 km/h
Convert km/h to m/s
(41×1000) / 3600
Speed = 11.39 m/s
Using speed formula to calculate time
Speed = distance/time
11.39 = 61/t
t = 61/11.39
t = 5.36s
Where the distance = 61 m
Maximum acceleration = velocity /time
Maximum acceleration = 11.39/5.36
Maximum acceleration = 2.13 m/s^2
The maximum acceleration value occurs when the sprinter is starting from rest and attaining the maximum speed.
A column of soldiers, marching at 100 steps per minute, keep in step with the beat of a drummer at the head of the column. It is observed that the soldiers in the rear end of the column are striding forward with the left foot when the drummer is advancing with the right. What is the approximate length of the column? (Take the speed of sound to be 343 m/s.)
Answer:
The value is [tex]D = 205.8 \ m [/tex]
Explanation:
The time taken for the column to take a step mathematically represented as
100 steps => 1 minutes => 60 seconds
1 step => t
=> [tex]t = 0.6 \ s [/tex]
Generally the length of the column is mathematically represented as
[tex]D = v * t[/tex]
substituting 343 m/s for v we have
[tex]D = 343 * 0.6 [/tex]
=> [tex]D = 205.8 \ m [/tex]
what would the answer be ?
Answer:
im going between 2 of them b and c but i would have choose b
In electronic circuits it is not unusual to encounter currents in the microampere range. Assume a 36 μA current, due to the flow of electrons. Part A What is the average number of electrons per second that flow past a fixed reference cross section that is perpendicular to the direction of flow?
Answer:
n = 2.25 x 10¹⁴ electrons/s
Explanation:
The amount of electric current is defined as the electric charge passing through an area per unit time. Hence:
I = q/t
where,
I = Current
q = amount of charge
t = time interval
but,
q = ne
therefore,
I = ne/t
where,
n = no. of electrons
e = charge on single electron = 1.6 x 10⁻¹⁹ C
t = 1 s (for electrons passing per second)
I = Current = 36 μA = 3.6 x 10⁻⁵ A
Therefore,
3.6 x 10⁻⁵ A = n(1.6 x 10⁻¹⁹ C)/1 s
n = (3.6 x 10⁻⁵ A)/(1.6 x 10⁻¹⁹ C)
n = 2.25 x 10¹⁴ electrons/s
3) An explorer walks 13 km due east, then 18 km north, and finally 3 km west.
a) What is the total distance walked?
b) What is the resulting displacement of the explorer from the starting point?
Answer: 34 km, 21 km 61 degrees north of east
Explanation: distance = 13 + 3 + 18 = 34
displacement = 13 - 3 = 10
10^2 + 18^2 = 424
find the square root of 424 ( 20.5 rounded to 21 )
The total distance walked is 34 km and the resulting displacement is 20.6 km.
a) The total distance is gotten by summing up all the distance.
Total distance = Distance moved east + distance moved north + distance moved west
Total distance = 13 km + 18 km + 3 km = 34 km
b) The displacement is the distance from the beginning point to end point.
Displacement² = 18² + (13 - 3)² = 18² + 10²
Displacement² = 424
Displacement = 20.6 km
Therefore the total distance walked is 34 km and the resulting displacement is 20.6 km.
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For any object in projectile motion, the vertical velocity is independent of gravity?
Answer:
the y axis, every moment is subjected to a vertical acceleration directed towards the center of the Earth, which is called the acceleration of gravity
Explanation:
In projectile launching, the movement is separated into two movements, one on the x-axis and the other on the y-axis, related through time.
In horizontal movement, the speed is constant, because there is no acceleration in this axis, the effect of air friction is almost always eliminated.
In the other movement on the y axis, every moment is subjected to a vertical acceleration directed towards the center of the Earth, which is called the acceleration of gravity.
This value of the eceleration of gravity is constant for small distances
compared to the radius of the Earth, for higher altitudes an expansion in beings of the distance is used, giving a linear dependence.
What is the total electric charge of 2.5 kg of (a) electrons and (b) protons?
Answer:
a
[tex]Q_e = -4.39 *10^{11} \ C [/tex]
b
[tex]Q_p = + 2.395*10^{-8} \ C [/tex]
Explanation:
Generally the number of electron in the given mass is mathematically evaluated as
[tex]N_e = \frac{2.5}{m_e }[/tex]
Here m_e is the mass of electron with value [tex]m_e = 9.11 * 10^{-31} \ kg[/tex]
=> [tex]N_e = \frac{2.5}{ 9.11 * 10^{-31} }[/tex]
=> [tex]N_e =2.74 *10^{30} \ electrons [/tex]
The total electric charge is mathematically represented as
[tex]Q_e = N_e * e[/tex]
Here e is the charge on a single electron with value [tex]e = 1.60 *10^{-19} \ C[/tex]
So
[tex]Q_e = -2.74 *10^{30} * 1.60 *10^{-19} [/tex]
[tex]Q_e = -4.39 *10^{11} \ C [/tex]
The negative sign is because we are considering electron
Generally the number of protons in the given mass is mathematically evaluated as
[tex]N_p = \frac{2.5}{m_p }[/tex]
Here m_p is the mass of electron with value [tex]m_e = 1.67 * 10^{-27} \ kg[/tex]
=> [tex]N_p = \frac{2.5}{ 1.67 * 10^{-27} }[/tex]
=> [tex]N_p =1.497 *10^{27} \ protons [/tex]
The total electric charge is mathematically represented as
[tex]Q_p = + N_p * e[/tex]
Here p is the charge on a single proton with value [tex]p = 1.60 *10^{-19} \ C[/tex]
So
[tex]Q_p = +1.497 *10^{27} * 1.60 *10^{-19} [/tex]
[tex]Q_p = + 2.395*10^{-8} \ C [/tex]
It is easier to open the lid of a can using a spoon why?
Answer:
It depends on how you use the spoon...
if you are keeping one end of the spoon and pressing other end, the force you provide is supported by the force due to gravity... Hence it is easy to open this way :)
F + G ... Where F is the force you provide and G is the force due to gravity.
URGENT URGENT DUE SOON!!
Which statement describes a question that can guide the design of a
scientific investigation?
A. It asks about a cause-and-effect relationship between two
variables.
B. It asks about the preferred outcome of the investigation.
C. It asks about whether a controlled variable is necessary.
D. It asks about how the observations will be organized.
Answer:
A.
Explanation:
Im just guessing because it sounds like it describes a hypothesis the most and you said its urgent
what does Aristotle say about the good life?Does it still stand contemporary world?
Answer:
Explanation:
Aristotle was a philosopher and a scientist who believed good life for humans is ultimately about happiness and this happiness can be achieved by virtue or high moral standards.
In the contemporary world, generally, everyone's ultimate goal is still to be happy but achieving this feat by being virtuous or having a high moral standard is fading away in our contemporary world as people just want to be happy by any means possible. Some amass wealth illegally just to be happy while others do things that are considered immoral (by the society) to be happy hence people still want a good life but not a way described by Aristotle.
Aristotle stated that a good life is about happiness and this can only be achieved through high moral standards.
The statement that a good life is about happiness and this can only be achieved through high moral standards doesn't stand anymore in the contemporary world. People now engage in immoral things as long as it brings them happiness. Even though the ultimate goal of people is to be happy, there are different ways that people go about in achieving this. There are some people who get their wealth illegally just to ensure that they are happy. This source of happiness is gotten through vices.In conclusion, Aristotle's statement about the good life is not applicable in the contemporary world anymore.
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A small block with mass 0.200 kg is released from rest at the top of a frictionless incline. The block travels a distance 0.394 m down the incline in 2.00 s. The 0.200 kg block is replaced by a 0.400 kg block. If the 0.400 kg block is released from rest at the top of the incline, how far down the incline does it travel in 2.00 s
Answer:
The distance traveled by the second block is 0.197 m
Explanation:
Given;
mass of the small block, m₁ = 0.2 kg
distance traveled by the block, d₁ =0.394 m
time of travel, t = 2 s
mass of the second block, m₂ = 0.4 kg
distance traveled by the second block, d₂ = ?
The work done per unit time on the inclined plane is given by;
[tex]\frac{f_1d_1}{t} = \frac{f_1d_1}{t}\\\\f_1d_1 = f_2d_2\\\\m_1gd_1 = m_2gd_2\\\\m_1d_1 = m_2d_2\\\\d_2 = \frac{m_1d_1}{m_2} \\\\d_2 = \frac{0.2*0.394}{0.4}\\\\d_2 = 0.197 \ m[/tex]
Therefore, the distance traveled by the second block is 0.197 m
Each insulated beaker contains equal amounts of the same fluid. The starting temperature of beaker A was 100.0 degrees Celsius and the starting temperature of beaker B was 0 degrees Celsius. At 5 minutes, the temperature of Beaker A was 82 and the temperature of Beaker B was 18. Assuming no heat was lost, what is the best estimate for the temperature of each beaker at 10 minutes?
Answer:
Correct Answer: B. Beaker A will be 72 °C and beaker B will be 28 °C.
This one is actually right!
half life of a given sample of radium is 22 years the sample will reduce to 25% of its original value after
Answer:
44 years
Explanation:
Use half life equation:
A = A₀ (½)^(t / T)
where A is the final amount,
A₀ is the initial amount,
t is time,
and T is the half life.
0.25 A₀ = A₀ (½)^(t / 22)
0.25 = (½)^(t / 22)
t / 22 = 2
t = 44
44 sec is half life of a given sample of radium is 22 years the sample will reduce to 25% of its original value.
What is half-life?The period of time it takes for one-half of a radioactive isotope to decay is known as the half-life. A given radioactive isotope's half-life is constant; it is unaffected by external factors and independent of the isotope's starting concentration.
Use half life equation:
A = A₀ (½)^(t / T)
where A is the final amount,
A₀ is the initial amount,
t is time,
and T is the half life.
0.25 A₀ = A₀ (½)^(t / 22)
0.25 = (½)^(t / 22)
t / 22 = 2
t = 44 sec
44 sec is half life of a given sample of radium is 22 years the sample will reduce to 25% of its original value.
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An unstrained horizontal spring has a length of 0.34 m and a spring constant of 180 N/m. Two small charged objects are attached to this spring, one at each end. The charges on the objects have equal magnitudes. Because of these charges, the spring stretches by 0.024 m relative to its unstrained length. Determine (a) the possible algebraic signs and (b) the magnitude of the charges. g
Answer:
a
[tex]q_1 = -1.389 *10^{-5} \ C [/tex] , [tex]q_2 = -1.389 *10^{-5} \ C [/tex]
OR
[tex]q_1 = 1.389 *10^{-5} \ C [/tex] , [tex]q_2 = 1.389 *10^{-5} \ C [/tex]
b
[tex]q_1 = 1.389 *10^{-5} \ C [/tex] and [tex]q_2 = 1.389 *10^{-5} \ C [/tex]
Explanation:
Generally the force exerted on the string is mathematically represented as
[tex]F = k * e[/tex]
substituting values 180 N/m for k and 0.024 m for e
[tex]F = 180 * 0.024[/tex]
[tex]F = 4.32 \ N[/tex]
This force can also equivalent to the electrostatic force between the charges i.e
[tex]F = k * \frac{q^2}{ r^2}[/tex]
substituting [tex]9*10^{9}\ kg\cdot m^3\cdot s^{-4} \cdot A^{-2}.[/tex] for k and ( 0.34 + 0.024 = 0.364 m) for r we have
[tex] 4.32= 9*10^{9} * \frac{q^2}{ (0.364)^2}[/tex]
[tex]q = \sqrt{1.929 *10^{-10}}[/tex]
[tex]q = 1.389 *10^{-5} \ C [/tex]
Given the spring was stretched it means that the force between the charges is a repulsive for which tell us that both charge are of the same sign thus the possible algebraic signs of the charges are
[tex]q_1 = -1.389 *10^{-5} \ C [/tex] , [tex]q_2 = -1.389 *10^{-5} \ C [/tex]
OR
[tex]q_1 = 1.389 *10^{-5} \ C [/tex] , [tex]q_2 = 1.389 *10^{-5} \ C [/tex]
Which of the following is the best thermal conductor?
A. Fiberglass.
B. Stainless.
C. Steel.
D. Wood.
E. Silver
Answer:
steel because being alloy in metal it has free electrons in metals
What was the current annual rate of inflation in 2009?
pls help !
Answer:
The 2009 inflation rate was -0.36%. The current year-over-year inflation rate (2019 to 2020) is now 0.99%
Explanation:
how many legs a cow has
What the differences between static and kinetic friction?
Answer:
Static friction prevents a stationary object from moving while kinetic or dynamic friction slows down a moving object.
Explanation:
Static Friction is the maximum force that must be overcome before a stationary object begins to move, while kinetic or dynamic friction is the maximum force that must be overcome for an object in motion to continue moving at a uniform velocity.
Static friction keeps a stationary object at rest, once the Force of Static friction is overcome, the Force of Kinetic friction is what slows down the moving object.
An oil pump is drawing 44kW while pumping oil with a density of 860 kg/m3 at a rate of 0.1 m3/s. The inlet and outlet diameters of the pipe are 8cm and 12cm, respectively. If the pressure increases by 500kPa going through the pump and the motoreffi ciency is 90%, determine the mechanical efficiency of the pump
Answer:
The mechanical efficiency of the pump is 91.8 %
Explanation:
Given;
input power, p = 44 kw
density of oil, ρ = 860 kg/m³
motor efficiency, η = 90 %
inlet diameter, d₁ = 8 cm
outlet diameter, d₂ = 12 cm
volume flow rate, V = 0.1 m³/s
pressure rise, P = 500kPa
output power = motor efficiency x input power
output power = 0.9 x 44 = 39.6 kW
Thus, the mechanical input power = 39.6 kW
The mechanical output power is given by change in mechanical energy;
[tex]E = mgh + \frac{m}{2} (v_2^2 - v_1^2) \\\\E = \rho V g h + \frac{\rho V}{2} [(\frac{V_2}{\pi r_2^2} )^2 - (\frac{V_1}{\pi r_1^2})^2]\\\\E = PV + \frac{\rho V^3}{2\pi^2} [\frac{1}{ r_2^4} - \frac{1}{ r_1^4}]\\\\E = (500 *10^3)(0.1) + \frac{(860)(0.1)^3}{2\pi^2} [\frac{1}{ 0.06^4} - \frac{1}{ 0.04^4}]\\\\E = 50000 -13653.51\\\\E = 36346.48 \ W\\\\E = 36.347 \ kW[/tex]
The mechanical efficiency is given by
η = mechanical output power / mechanical input power
η = 36.347 / 39.6
η = 0.918
η = 91.8 %
Therefore, the mechanical efficiency of the pump is 91.8 %
why is evidence good for practice
Answer:
HEYO BRO! (my bad for da caps)
Explanation:
Evidence-based practice (EBP) is the idea that occupational practices ought to be based on scientific evidence. That at first sight may seem to be obviously desirable, but the proposal has been controversial.
Happy to Help From, Adam